A limit problem about $a_{n+1}=a_n+\frac{n}{a_n}$ Let $a_{n+1}=a_n+\frac{n}{a_n}$ and $a_1>0$. Prove $\lim\limits_{n\to \infty} n(a_n-n)$ exists.
In my view, maybe we can use 
$${a_{n + 1}} = {a_n} + \frac{n}{{{a_n}}} \Rightarrow {a_{n + 1}} - \left( {n + 1} \right) = \left( {{a_n} - n} \right)\left( {1 - \frac{1}{{{a_n}}}} \right).$$
And then 
$${a_n} - n = \left( {{a_1} - 1} \right)\prod\limits_{k = 1}^{n - 1} {\left( {1 - \frac{1}{{{a_k}}}} \right)} .$$
By Stolz formula, we have 
\begin{align*}
&\mathop {\lim }\limits_{n \to \infty } n\left( {{a_n} - n} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{\frac{1}{{{a_n} - n}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\frac{1}{{{a_{n + 1}} - \left( {n + 1} \right)}} - \frac{1}{{{a_n} - n}}}}\\
 = &\mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{{a_n} - {a_{n + 1}} + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{ - \frac{n}{{{a_n}}} + 1}}\\
 = &\mathop {\lim }\limits_{n \to \infty } {a_n}\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right].
\end{align*}
And how can we continue?
 A: We have:
$$ a_n(a_{n+1}-a_n) = n $$
so:
$$ a_{n+1}^2-a_{n}^2 = a_{n+1}(a_{n+1}-a_n) + n = n\left(1+\frac{a_{n+1}}{a_n}\right)=2n+\frac{n^2}{a_n^2}$$
and:
$$ a_{N+1}^2-a_1^2 = N(N+1)+\sum_{n=1}^{N}\frac{n^2}{a_n^2} $$
from which $a_{N+1}\geq \sqrt{N(N+1)}$ and $a_n\geq \sqrt{(n-1)n}$. 
If we plug this inequality  back into the previous line, we get:
$$\begin{eqnarray*} a_{N+1}^2 &\leq& N(N+1)+a_1^2+\frac{1}{a_1^2}+\sum_{n=2}^{N}\frac{n}{n-1}\\&=& (N+1)^2+\left(a_1-\frac{1}{a_1}\right)^2+H_{N-1}.\end{eqnarray*} $$
The process continues by keep turning lower/upper bounds into tighter upper/lower bounds. 
Can you check it proves your statement?
A: Let's start with some observations:
1) You can clearly see from what you have developed that $a_n\geq n$ (equality for $a_1=1$. For the rest we assume $a_1=1+\alpha>1$).
2) It is also easy to see that $a_{n}-n$ is decreasing. So if $a_1=\alpha+1$ then $a_n-n<\alpha$.
3) Consider the  product $\prod\limits_{k = 1}^{n - 1} {\left( {1 - \frac{1}{{{a_k}}}} \right)} $, we can show it converges to zero using following inequalities:
$$
\sum\limits_{k = 1}^{n - 1}\log {\left( {1 - \frac{1}{{{a_k}}}} \right)} \leq \sum\limits_{k = 1}^{n - 1}({ - \frac{1}{{{a_k}}}}) \leq \sum\limits_{k = 1}^{n - 1} \frac{-1}{k+\alpha}\to-\infty
$$
Hence $a_n-n\to 0$.
4) Define $c_n=a_n-n$ and $b_n=nc_n$; we have:
$$
b_{n+1}-b_n=c_n(\frac{c_n}{n+c_n}-\frac{1}{n+c_n}).
$$
The term on the right goes to zero and therefore $b_{n+1}-b_n\to 0$.
